Exercises — State and apply Ohm's Law (V = IR)

We want $I$, and we know $V$ and $R$. Cover $I$ in the VIR triangle (V on top, I and R on the bottom) → what remains is $V$ over $R$. $\boxed{I=\frac{V}{R}}$ This is just $V=IR$ divided by $R$ on both sides. Nothing measured, nothing converted — pure recognition.

• Current $I$ → ampere (A) — how much charge flows per second (see Electric Current).
• Potential difference $V$ → volt (V) — energy per unit charge (see Potential Difference (Voltage)).
• Resistance $R$ → ohm (Ω) — opposition to flow (see Resistance).

Want $I$, know $V$ and $R$: $I=\frac{V}{R}=\frac{9}{3}=3\text{A}$

Convert first — the formula needs base SI units: $150\text{ mA}=0.150\text{ A},4.7\text{ k}\Omega =4700\Omega$ Then $V=IR=0.150\times 4700=705\text{V}$ Shortcut check: mA × kΩ = V directly, so $150\times 4.7=705\text{ V}$. ✓ (The ×1000 and ÷1000 cancel.)

Convert the current: $50\mu \text{A}=50\times 10^{-6}\text{ A}=0.00005\text{ A}$. $R=\frac{V}{I}=\frac{2.5}{0.00005}=50000\Omega =50\text{ k}\Omega$ A tiny current for a decent voltage → large resistance. That matches intuition.

On this graph, gradient $=V/I=R$. Both lines pass through the origin (ohmic), so we can read $R$ off any single point. $R_A=\frac{4}{2}=2\Omega ,R_B=\frac{8}{2}=4\Omega$ B has the greater resistance, twice A's. Look at the picture: the steeper orange line is the bigger resistor — steeper means "more volts needed per amp," i.e. more opposition.

Predict: $R$ is constant, and $V=IR$ means $I\propto V$. Voltage went ×3 ($4\to 12$), so current goes ×3. Old current: $I_1=4/2=2\text{ A}$ → predicted new current $=6\text{ A}$. Verify: $I_2=V/R=12/2=6\text{ A}$. ✓ Matches — that is the "directly proportional" claim in action.

Compute $R=V/I$ at each point: $\frac{2}{1}=2\Omega ,\frac{6}{2}=3\Omega ,\frac{12}{3}=4\Omega$ The ratio changes, so $R$ is not constant → the component is non-ohmic (its V–I graph curves upward, like a filament lamp). Ohmic means one fixed gradient through the origin.

(a) In series the same current flows through both, and resistances add: $R_{\text{total}}=R_1+R_2=4+8=12\Omega$ (b) Apply Ohm's Law to the whole circuit: $I=\frac{V}{R_{\text{total}}}=\frac{12}{12}=1\text{A}$ (c) Same $1\text{ A}$ flows through each; apply $V=IR$ to each individually: $V_1=I{R}_1=1\times 4=4\text{ V},V_2=I{R}_2=1\times 8=8\text{ V}$ Check: $V_1+V_2=4+8=12\text{ V}$ = battery voltage. ✓ The voltages share out in proportion to resistance.

(a) In parallel, each branch feels the full $12\text{ V}$. Apply Ohm's Law to each branch: $I_1=\frac{V}{R_1}=\frac{12}{4}=3\text{ A},I_2=\frac{V}{R_2}=\frac{12}{8}=1.5\text{ A}$ (b) Currents add at the junction: $I_{\text{total}}=I_1+I_2=3+1.5=4.5\text{ A}$ (c) Equivalent resistance from the whole circuit: $R_{\text{eq}}=\frac{V}{I_{\text{total}}}=\frac{12}{4.5}=2.667\Omega \approx 2.67\Omega$ Notice $R_{\text{eq}}$ is smaller than either resistor — parallel paths give current more ways through, i.e. less overall opposition.

First get the voltage across it: $V=IR=2\times 6=12\text{ V}$. Then power: $P=VI=12\times 2=24\text{W}$ Equivalently $P=I^{2}R=2^2\times 6=24\text{ W}$. Both routes agree.

(a) $R=V/I$ still defines the resistance at each instant: $R_{\text{cold}}=\frac{2}{0.5}=4\Omega ,R_{\text{hot}}=\frac{12}{2}=6\Omega$ (b) More voltage → more current → more electrical energy dumped into the filament → it heats up. Hotter metal atoms vibrate more, so drifting electrons collide with them more often. More collisions = more opposition = higher $R$ (see Non-ohmic Components). (c) No. The resistance rose from $4\Omega$ to $6\Omega$, so there is no single fixed value. The V–I graph curves; the lamp is non-ohmic. $V=IR$ is not a broken equation here — it still defines $R$ at each point — but $R$ is no longer a constant, so you cannot use one measurement to predict another.

Series → total resistance adds, then Ohm's Law on the whole loop. Cold: $R_{\text{total}}=10+50=60\Omega$, so $I_{\text{cold}}=\frac{12}{60}=0.2\text{ A}$ Warm: $R_{\text{total}}=10+10=20\Omega$, so $I_{\text{warm}}=\frac{12}{20}=0.6\text{ A}$ The thermistor's resistance falls when heated (opposite to a metal), so total resistance drops and the current rises from $0.2\text{ A}$ to $0.6\text{ A}$. This is exactly how a thermistor acts as a temperature sensor.

We need a link between power, voltage and resistance. Combine $P=VI$ with $I=V/R$: $P=V\cdot \frac{V}{R}=\frac{V^2}{R}$ Why this form? We're given $P$ and $R$ and want $V$, so we eliminate $I$. Rearrange for $V$: $V=\sqrt{PR}=\sqrt{0.25\times 100}=\sqrt{25}=5\text{V}$ So keep the voltage at or below $5\text{ V}$. Check the current at that limit: $I=V/R=5/100=0.05\text{ A}$, and $P=VI=5\times 0.05=0.25\text{ W}$. ✓ Exactly at the rating.

Series shares current (voltages split). Parallel shares voltage (currents split).

Smaller — extra parallel paths reduce total opposition.

$V$ from $P$ and $R$ in one step? $V=\sqrt{PR}$, from $P=V^2/R$.

It is non-ohmic; heating raises $R$, so the cold value doesn't apply at the hot operating point.